\(PT\Leftrightarrow x^2=\dfrac{y^2-1}{y+2}=\dfrac{y^2-4+3}{y+2}=y-2+\dfrac{3}{y+2}\)
Vì \(x,y\in Z\)\(\Rightarrow\dfrac{3}{y+2}\in Z\)
\(y+2\in\left\{-1;1;-3;3\right\}\)
\(\Rightarrow y\in\left\{-3;-1;-5;1\right\}\)
Thử lại tìm x, ta thu được (x,y)=(0;-1);(0;1)